Pt(s)|H₂(g)(1 bar)|H + (aq)(1 M)||M³⁺(aq),M + (aq)|Pt(s)
The E_(cell) for the given cell is 0.1115 V at 298 K when [ M + (aq) ]/[ M³⁺(aq) ] = 10ᵃ
The value of a is ____________
Given : E_(M³⁺/M +)^θ = 0.2 V
2.303RT/F = 0.059V

Solution

Chemistry ma’am · 4 steps

Hydrogen is the reference here, so E°_(cell) is just 0.2 V. Find n, then let Nernst do the rest.

1. Write the reactions. Anode: H₂ → 2H⁺ + 2e⁻. Cathode: M³⁺ + 2e⁻ → M⁺. So n = 2, and E°_(cell) = 0.2 - 0 = 0.2 V.
2. Reaction quotient. For H₂ + M³⁺ → 2H⁺ + M⁺, Q = [H⁺]²[M⁺]/p_(H₂)[M³⁺]. With [H⁺] = 1 M and p_(H₂) = 1 bar, this is [M⁺]/[M³⁺] = 10ᵃ.

   E_(cell) = E°_(cell) - 0.059/n log Q ⇒ 0.1115 = 0.2 - 0.059/2 a

3. Solve for a. 0.0295 a = 0.2 - 0.1115 = 0.0885, so a = 0.0885/0.0295 = 3.
4. So. The ratio [M⁺]/[M³⁺] = 10³, and a = 3.

The common slip: 1.5. Taking n = 1 makes the step 0.059 a = 0.0885, which gives 1.5 instead of 3.

The idea to keep. Always read n from the balanced half-reaction, not from a charge. M³⁺ → M⁺ is a 2-electron change, and that 2 sits under 0.059 in the Nernst equation.

Answer from the NTA final answer key. Solution by PrepWiser.